What type of matter is gravel

5.) During the electrolysis of dilute copper(II)sulphate solution using carbon 1 point

Neporo4naja [7]

Answer:

A g I need the answer too

Explanation:

6 0

3 years ago

How to make 0.01 Molar solution in 30 ml of water

Leviafan [203]

Answer:

You see the concentration decreased ten fold, from 1 M to 0.1 M.

You can simply increase the volume by ten times.

For example

If you have 250 mL of 1 molar HCl, you can add distilled water upto 2500 mL. Now the concentration is 0.1 molar.

If you want to use the serial dilution method, you can use the C1V1=C2V2 equation.

C1 = starting concentration.

C2 = final concentration.

V is for volume.

So let's say you have 500 mL of 1 M HCl and you only want 50 mL of 0.1 M HCl.

1 M x V1 = 0.1 M x 50 mL

V1 = 5 mL

So you take 5 mL of the original (stock) solution and dilute it up to 50 mL. Then you have 50 mL of 0.1 M HCl.

Explanation:

4 0

3 years ago

What is the freezing point of a solution of 0.500 mol of sucros in 500.0 mL of water?

andrew11 [14]

The freezing point would be 3.72 *C

8 0

3 years ago

Balance the following redox equations by the half-reaction method:

frozen [14]

Answer:

a) Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O → 2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2 → MnO2 + 2H2O + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

(b) Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution)

Step 1: The half reactions

Reduction: Bi(OH)3 + 3e- → Bi

Oxidation : Sno2^2- → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e- → Bi + 3OH-)

3* (Sno2^2- +2OH- → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e- → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e- +14H+ → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

(d) ClO3^- + Cl^− → Cl^2 + ClO^2 (in acidic solution)

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

ClO3^- + e- → ClO2

2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

(e) Mn^2 + BiO3^− → Bi^3 + MnO^4− (in acidic solution)

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

6 0

4 years ago

Pls how do you test for the hydrogen ion concentration of a bottle of water.

iren [92.7K]

Using the pH=-log[ hydrogen ions], assuming the pH of water is 7 then using the concept that kw=[ OH] [H+], then the [H+]= 10 power minus 7,since kw=10 to the power minus 14

4 0

4 years ago